判断是否为二分图
算法：
- 选取一个未染色的点u进行染色
- 遍历u的相邻节点v：若v未染色，则染色成与u不同的颜色，并对v重复第2步；若v已经染色，如果 u和v颜色相同，判定不可行退出遍历。
- 若所有节点均已染色，则判定可行。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<string> 6 #include<queue> 7 #include<algorithm> 8 #include<map> 9 #include<iomanip> 10 #include<climits> 11 #include<string.h> 12 #include<cmath> 13 #include<stdlib.h> 14 #include<vector> 15 #include<stack> 16 using namespace std; 17 #define INF 1000000007 18 #define MAXN 40010 19 #define Mod 1000007 20 #define N 10007 21 #define NN 30 22 #define sigma_size 3 23 const int maxn = 6e5 + 10; 24 using namespace std; 25 typedef long long LL; 26 27 bool vis[10010]; 28 bool col[10010]; 29 vector<int> G[10010]; 30 int n, m; 31 int u, v; 32 33 bool dfs(int u, bool c) 34 { 35 if (vis[u]) 36 return col[u] == c; 37 else { 38 vis[u] = true; 39 col[u] = c; 40 for (int i = 0; i < G[u].size(); ++i) { 41 int v = G[u][i]; 42 if (!dfs(v,!c)) return false; 43 } 44 return true; 45 } 46 } 47 48 void process() 49 { 50 memset(vis,0,sizeof(vis)); 51 memset(col,0,sizeof(col)); 52 for (int i = 0; i < n; ++i) 53 G[i].clear(); 54 cin >> n >> m; 55 for (int i = 0; i < m; ++i) { 56 cin >> u >> v; 57 u--, v--; 58 G[u].push_back(v); 59 G[v].push_back(u); 60 } 61 bool re = 1; 62 for (int i = 0; i < n; ++i) 63 if (!vis[i]) 64 re &= dfs(i, 0); 65 if (re) puts("Correct"); 66 else puts("Wrong"); 67 } 68 69 int main() 70 { 71 int T; 72 cin >> T; 73 while (T--) 74 process(); 75 //system("pause"); 76 return 0; 77 }
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<string> 6 #include<queue> 7 #include<algorithm> 8 #include<map> 9 #include<iomanip> 10 #include<climits> 11 #include<string.h> 12 #include<cmath> 13 #include<stdlib.h> 14 #include<vector> 15 #include<stack> 16 #include<set> 17 using namespace std; 18 #define INF 1000000007 19 #define MAXN 40010 20 #define Mod 1000007 21 #define N 10007 22 #define NN 30 23 #define sigma_size 3 24 const int maxn = 6e5 + 10; 25 using namespace std; 26 typedef long long LL; 27 28 int n, m; 29 int u, v; 30 int color[10010]; 31 vector<int> G[40010]; 32 33 bool bipartite(int u) 34 { 35 for (int i = 0; i < G[u].size(); ++i) { 36 int v = G[u][i]; 37 if (color[v] == color[u]) return false; 38 if (!color[v]) { 39 color[v] = 3 - color[u]; 40 if (!bipartite(v)) return false; 41 } 42 } 43 return true; 44 } 45 46 void process() 47 { 48 cin >> n >> m; 49 memset(color, 0, sizeof(color)); 50 for (int i = 0; i <= n; ++i) 51 G[i].clear(); 52 for (int i = 0; i < m; ++i) { 53 cin >> u >> v; 54 G[u].push_back(v); 55 G[v].push_back(u); 56 } 57 bool re = true; 58 for (int i = 1; i <= n; ++i) 59 if (!color[i]) { 60 color[i] = 1; 61 re &= bipartite(i); 62 } 63 if (re) puts("Correct"); 64 else puts("Wrong"); 65 } 66 67 int main() 68 { 69 int T; 70 cin >> T; 71 while (T--) 72 process(); 73 //system("pause"); 74 return 0; 75 }